Union with Empty Set

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Theorem

The union of any set with the empty set is the set itself:

$S \cup \O = S$


Proof 1

\(\displaystyle S\) \(\subseteq\) \(\displaystyle S\) $\quad$ Set is Subset of Itself $\quad$
\(\displaystyle \O\) \(\subseteq\) \(\displaystyle S\) $\quad$ Empty Set is Subset of All Sets $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle S \cup \O\) \(\subseteq\) \(\displaystyle S\) $\quad$ Union is Smallest Superset $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle S\) \(\subseteq\) \(\displaystyle S \cup \O\) $\quad$ Set is Subset of Union $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle S \cup \O\) \(=\) \(\displaystyle S\) $\quad$ Definition of Set Equality $\quad$

$\blacksquare$


Proof 2

From Empty Set is Subset of All Sets:

$\O \subseteq S$

From Union with Superset is Superset‎:

$S \cup \O = S$

$\blacksquare$


Also see


Sources